Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		double[] p = new double[n];
		sa = br.readLine().split(" ");
		for (int i = 0; i < n; i++) {
			p[i] = Double.parseDouble(sa[i]);
		}
		br.close();

		double[][] dp = new double[n + 1][n + 1];
		dp[0][0] = 1;
		for (int i = 1; i <= n; i++) {
			double pi = p[i - 1];
			dp[i][0] = dp[i - 1][0] * (1 - pi);
			for (int j = 1; j <= i; j++) {
				dp[i][j] = dp[i - 1][j - 1] * pi
						+ dp[i - 1][j] * (1 - pi);
			}
		}

		double ans = 0;
		for (int i = n / 2 + 1; i <= n; i++) {
			ans += dp[n][i];
		}
		System.out.println(ans);
	}
}